Topological Symmetry Groups of Small Complete Graphs

Abstract

Topological symmetry groups were originally introduced to study the symmetries of non-rigid molecules, but have since been used to study the symmetries of any graph embedded in ℝ³. In this paper, we determine for each complete graph K_n with n ≤ 6, what groups can occur as topological symmetry groups or orientation preserving topological symmetry groups of some embedding of the graph in ℝ³.

1. Introduction

Molecular symmetries are important in many areas of chemistry. Symmetry is used in interpreting results in crystallography, spectroscopy, and quantum chemistry, as well as in analyzing the electron structure of a molecule. Symmetry is also used in designing new pharmaceutical products. But what is meant by a “symmetry” depends on the rigidity of the molecule in question.

For rigid molecules, the group of rotations, reflections, and combinations of rotations and reflections, is an effective way of representing molecular symmetries. This group is known as the point group, of the molecule because it fixes a point of ℝ³. However, some molecules can rotate around particular bonds, and large molecules can even be somewhat flexible. For example, supramolecular structures constructed through self-assembly may be somewhat conformationally flexible. Even relatively small molecules may contain rigid molecular subparts that rotate on hinges around particular bonds. For example, the left and right sides of the biphenyl derivative illustrated in Figure 1 rotate simultaneously, independent of the central part of the molecule. Because of these rotating pieces, this molecule is achiral though it cannot be rigidly superimposed on its mirror form. A detailed discussion of the achirality of this molecule can be found in [1].

In general, the amount of rigidity of a given molecule depends on its chemistry not just its geometry. Thus a purely mathematical definition of molecular symmetries that accurately reflects the behavior of all molecules is impossible. However, for non-rigid molecules, a topological approach to classifying symmetries including achirality can add important information beyond what is obtained from the point group. Such an approach could be useful to the study of supramolecular chirality, since structures constructed through self-assembly may be large and somewhat flexible or contain subparts that can rotate around covalent or non-convalent bonds.

The topological symmetry group was first introduced by Jon Simon in 1987 in order to classify the symmetries of non-rigid molecules [2]. By comparing the topological symmetry group and the orientation preserving topological symmetry group of a particular structure, one can see whether the structure is achiral and if so, understand how its achirality fits together with its other topological symmetries.

In this paper, we determine both the topological symmetry groups and the orientation preserving topological symmetry groups of structures whose underlying form is that of a complete graph with no more than six vertices. A complete graph, K_n, is defined to be a graph with n vertices which has an edge between every pair of vertices. In Figure 2 we illustrate embeddings of the complete graphs K₃, K₄, and K₅. The class of complete graphs is an interesting class to consider because the automorphism group of K_n is the symmetric group S_n, which is the largest automorphism group of any graph with n vertices. For small values of n, there exist molecules whose underlying topological structure has the form of K_n. For example, a tetrahedral supramolecular cluster has the underlying structure of the complete graph K₄. If such a cluster contains a central atom which is bonded to the four corners of the tetrahedron, then the structure has the form of the complete graph K₅ (as illustrated on the right in Figure 2).

2. Background and Terminology

Though it may seem strange from the point of view of a chemist, the study of symmetries of embedded graphs is more convenient to carry out in the 3-dimensional sphere S³ = ℝ³ ∪ {∞} rather than in Euclidean 3-space, ℝ³. In particular, in ℝ³ every rigid motion is a rotation, reflection, translation, or a combination of these operations. Whereas, in S³ glide rotations provide an additional type of rigid motion. While a topological approach to the study of symmetries does not require us to focus on rigid motions, for the purpose of illustration it is preferable to display rigid motions rather than isotopies whenever possible. Thus throughout the paper we work in S³ rather than in ℝ³.

Definition 1

The topological symmetry group of a graph Γ embedded in S3 is the subgroup of the automorphism group of the graph, Aut(Γ), induced by homeomorphisms of the pair (S³, Γ). The orientation preserving topological symmetry group, TSG₊(Γ), is the subgroup of Aut(Γ) induced by orientation preserving homeomorphisms of (S³, Γ).

It should be noted that for any homeomorphism h of (S³, Γ), there is a homeomorphism g of (S³, Γ) which fixes a point p not on Γ such that g and h induce the same automorphism on Γ. By choosing p to be the point at ∞, we can restrict g to a homeomorphism of (ℝ³, Γ). On the other hand if we start with an embedded graph Γ in ℝ³ and a homeomorphism g of (ℝ³, Γ), we can consider Γ to be embedded in S³ = ℝ³ ∪ {∞} and extend g to a homeomorphism of S³ simply by fixing the point at ∞. It follows that the topological symmetry group of Γ in S³ is the same as the topological symmetry group of Γ in ℝ³. Thus we lose no information by working with graphs in S³ rather than graphs in ℝ³.

It was shown in [3] that the set of orientation preserving topological symmetry groups of 3-connected graphs embedded in S³ is the same up to isomorphism as the set of finite subgroups of the group of orientation preserving diffeomorphisms of S³, Diff₊(S³). However, even for a 3-connected embedded graph Γ, the automorphisms in TSG(Γ) are not necessarily induced by finite order homeomorphisms of (S³, Γ).

For example, consider the embedded 3-connected graph Γ illustrated in Figure 3. The automorphism (153426) is induced by a homeomorphism that slithers the graph along itself while interchanging the inner and outer knots in the graph. On the other hand, the automorphism (153426) cannot be induced by a finite order homeomorphism of S³ because there is no order three homeomorphism of S³ taking a figure eight knot to itself [4,5] and the embedded graph in Figure 3 cannot be pointwise fixed by a finite order homeomorphism of S³ [6].

On the other hand, Flapan proved the following theorem which we will make use of later in the paper.

Finite Order Theorem

[7] Let φ be a non-trivial automorphism of a 3-connected graph γ which is induced by a homeomorphism h of (S³, Γ) for some embedding Γ ofγ in S³. Then for some embedding Γ of γ in S³, the automorphism φ is induced by a finite order homeomorphism, f of (S³, Γ), and f is orientation reversing if and only ifh is orientation reversing.

In the definition of the topological symmetry group, we start with a particular embedding Γ of a graph γ in S³ and then determine the subgroup of the automorphism group of γ which is induced by homeomorphisms of (S³,Γ). However, sometimes it is more convenient to consider all possible subgroups of the automorphism group of an abstract graph, and ask which of these subgroups can be the topological symmetry group or orientation preserving topological symmetry group of some embedding of the graph in S³. The following definition gives us the terminology to talk about topological symmetry groups from this point of view.

Definition 2

An automorphism f of an abstract graph, γ, is said to be realizable if there exists an embedding Γ of γ in S³ such that f is induced by a homeomorphism of (S³, Γ). A group G is said to be realizable for γ if there exists an embedding Γ of γ in S³ such that TSG(Γ) = G. If there exists an embedding Γ such that TSG₊(Γ) = G, then we say G is positively realizable for γ.

It is natural to ask whether every finite group is realizable. In fact, it was shown in [3] that the alternating group A_m is realizable for some graph if and only if m < 5. Furthermore, in [8] it was shown that for every closed, connected, orientable, irreducible 3-manifold M, there exists an alternating group A_m which is not isomorphic to the topological symmetry group of any graph embedded in M.

3. Topological Symmetry Groups of Compete Graphs

For the special class of complete graphs K_n embedded in S³, Flapan, Naimi, and Tamvakis obtained the following result.

Complete Graph Theorem

[9] A finite group H is isomorphic to TSG₊(Γ) for some embedding Γ of a complete graph in S³ if and only if H is a finite subgroup of SO(3) or a subgroup of D_m × D_mfor some odd m.

This left open the question of what topological symmetry groups and orientation preserving topological symmetry groups are possible for embeddings of a particular complete graph K_n in S³. For each n > 6, this question was answered for orientation preserving topological symmetry groups in the series of papers [10-13]. These papers make use of a result that for n > 6, only a few types of automorphisms of K_n are realizable [7]. There are no comparable results available for automorphisms of K_n when n ≤ 6.

In the current paper, we determine which groups are realizable and which groups are positively realizable for each K_n with n < 6. This is the first family of graphs for which both the realizable and the positively realizable groups have been determined.

For n ≤ 3, this question is easy to answer. In particular, since K₁ is a single vertex, the only realizable or positively realizable group is the trivial group. Since K₂ is a single edge, the only realizable or positively realizable group is ℤ₂.

For n = 3, we know that Aut(K₃) ≅ S₃ ≅ D₃, and hence every realizable or positively realizable group for K₃ must be a subgroup of D₃. Note that for any embedding of K₃ in S³, the graph can be “slithered” along itself to obtain an automorphism of order 3 which is induced by an orientation preserving homeomorphism. Thus the topological symmetry group and orientation preserving topological symmetry group of any embedding of K₃ will contain an element of order 3. Thus neither the trivial group nor Z₂ is realizable or positively realizable for K₃. If Γ is a planar embedding of K₃ in S³, then TSG(Γ) = TSG₊(Γ) ≅ D₃. Recall that the trefoil knot 31 is chiral while the knot 81₇ is negative achiral and non-invertible. Thus if Γ is the knot 8₁₇, then no orientation preserving homeomorphism of (S³, Γ) inverts Γ, but there is an orientation reversing homeomorphism of (S³, Γ) which inverts Γ. Whereas, if Γ is the knot 8₁₇#3₁, then there is no homeomorphism of (S³, Γ) which inverts Γ.

Table 1. Realizable and positively realizable groups for K₃.

Embedding	TSG(Γ)	TSG+(Γ)
Planar	D3	D3
817	D3	ℤ3
817 #31	ℤ3	ℤ3

summarizes our results for K₃.

Determining which groups are realizable and positively realizable for K₄, K₅, and K₆ is the main point of this paper. In each case, we will first determine the positively realizable groups and then use the fact that either TSG₊(Γ) = TSG(Γ) or TSG₊(Γ) is a normal subgroup of TSG(Γ) of index 2 to help us determine the realizable groups.

4. Topological Symmetry Groups of K₄

In addition to the Complete Graph Theorem given above, we will make use of the following results in our analysis of positively realizable groups for K_n with n ≥ 4.

A₄ Theorem

[11] A complete graph K_m with m ≥ 4 has an embedding Γ in S³ such that TSG₊(Γ) ≅ A₄ if and only ifm ≡ 0, 1, 4, 5, 8 (mod 12).

A₅ Theorem

[11] A complete graph K_m with m ≥ 4 has an embedding Γ in S³ such that TSG₊(Γ) ≅ A₅ if and only ifm ≡ 0, 1, 5, 20 (mod 60).

S₄ Theorem

[11] A complete graph K_m with m ≥ 4 has an embedding Γ in S³ such that TSG₊(Γ) ≅ S₄ if and only if m ≡ 0, 4, 8, 12, 20 (mod 24).

Subgroup Theorem

[12] Let Γ be an embedding of a 3-connected graph γ in S³ with an edge that is not poinwise fixed by any non-trivial element of TSG₊(Γ). Then every subgroup of TSG₊(Γ) is positively realizable for γ.

It was shown in [12] that adding a local knot to an edge of a 3-connected graph is well-defined and that any homeomorphism of S³ taking the graph to itself must take an edge with a given knot to an edge with the same knot. Furthermore, any orientation preserving homeomorphism of S³ taking the graph to itself must take an edge with a given non-invertible knot to an edge with the same knot oriented in the same way Thus for n > 3, adding a distinct knot to each edge of an embedding of K_n in S³ will create an embedding Δ where TSG(Δ) and TSG₊(Δ) are both trivial. Hence we do not include the trivial group in our list of realizable and positively realizable groups for K_n when n > 3.

Finally, observe that for n > 3, for a given embedding Γ of K_n we can add identical chiral knots (whose mirror image do not occur in Γ) to every edge of Γ to get an embedding Γ such that TSG(Γ′) = TSG₊(Γ). Thus every group which is positively realizable for K_n is also realizable for K_n. We will use this observation in the rest of our analysis.

The following is a complete list of all the non-trivial subgroups of Aut(K₄) ≅ S₄ up to isomorphism: S₄, A₄, D₄, D₃, D₂, ℤ₄, ℤ₃, ℤ₂.

We will show that all of these groups are positively realizable, and hence all of the groups will also be realizable. First consider the embedding Γ of K₄ illustrated in Figure 4. The square $\overline{1234}$ must go to itself under any homeomorphism of (S³, Γ). Hence TSG₊(Γ) is a subgroup of D₄. In order to obtain the automorphism (1234), we rotate the square $\overline{1234}$ clockwise by 90° and pull $\overline{24}$ under $\overline{13}$. We can obtain the transposition (13) by first rotating the figure by 180° about the axis which contains vertices 2 and 4 and then pulling $\overline{13}$ under $\overline{24}$. Thus TSG₊(Γ) ≅ D₄. Furthermore, since the edge $\overline{12}$ is not pointwise fixed by any non-trivial element of TSG₊(Γ), by the Subgroup Theorem the groups ℤ₄, D₂ and ℤ₂ are each positively realizable for K₄.

Next, consider the embedding, Γ of K₄ illustrated in Figure 5. All homeomorphisms of (S³, Γ) fix vertex 4. Hence TSG₊(Γ) is a subgroup of D₃. The automorphism (123) is induced by a rotation, and the automorphism (12) is induced by turning the figure upside down and then pushing vertex 4 back up through the centre of 123. Thus TSG₊(Γ) = D₃. Since the edge 12 is not pointwise fixed by any non-trivial element of TSG₊(Γ), by the Subgroup Theorem, the group Z₃ is also positively realizable for K₄.

Thus every subgroup of Aut(K₄) is positively realizable. Now by adding appropriate equivalent chiral knots to each edge, all subgroups of Aut(K₄) are also realizable. We summarize our results for K₄ in

Table 2. Non-trivial realizable and positively realizable groups for K₄.

Subgroup	Realizable/Positively Realizable	Reason
S4	Yes	By S4 Theorem
A4	Yes	By A4 Theorem
D4	Yes	By Figure 4
D3	Yes	By Figure 5
D2	Yes	By Subgroup Theorem
ℤ4	Yes	By Subgroup Theorem
ℤ3	Yes	By Subgroup Theorem
ℤ2	Yes	By Subgroup Theorem

.

5. Topological Symmetry Groups of K₅

The following is a complete list of all the non-trivial subgroups of Aut(K₅) ≅ S₅:

• S₅, A₅, S₄, A₄, ℤ₅ × ℤ₄, D₆, D₅, D₄, D₃, D₂, ℤ₆, ℤ₅, ℤ₄, ℤ₃, ℤ₂ (see [14] and [15]).

The lemma below follows immediately from the Finite Order Theorem [7] (stated in the introduction) together with Smith Theory [6].

Lemma 1

Let n > 3 and let φ be a non-trivial automorphism of K_n which is induced by a homeomorphism h of (S³, Γ) for some embedding Γ of K_n in S³. If h is orientation reversing, then φ fixes at most 4 vertices. Ifh is orientation preserving, then φ fixes at most 3 vertices, and if φ has even order, then φ fixes at most 2 vertices.

We now prove the following lemma.

Lemma 2

Let n > 3 and let Γ be an embedding of K_n in S³ such that TSG₊(Γ) contains an element φ of even order m > 2. Then φ does not fix any vertex or interchange any pair of vertices.

Proof

By the Finite Order Theorem, K_n can be re-embedded as Γ so that φ is induced on Γ by a finite order orientation preserving homeomorphism h of (S³, Γ′). Suppose that φ fixes a vertex or interchanges a pair of vertices of Γ. Then fix(h) is non-empty, and hence by Smith Theory, fix(h) ≅ S¹. Let r = m/2. Then h^r induces an involution on the vertices of Γ, and this involution can be written as a product (a₁b₁) … (a_qb_q) of disjoint transpositions of vertices. Now for each i, h^r fixes a point on the edge a_ib. But fix(h^r) contains fix(h) and thus by Smith Theory fix(h^r) = fix(h). Hence h fixes a point on each edge a_ib_i. Thus h induces also (a₁ b₁) … (a_qb_q) on the vertices of Γ′. But this contradicts the hypothesis that the order of φ is m > 2.

By Lemma 2, there is no embedding of K₅ in S³ such that TSG₊(Γ) contains an element of order 4 or of order 6. It follows that TSG₊(Γ) cannot be D₆, ℤ₆, D₄ or ℤ₄.

Consider the embedding Γ of K₅ illustrated in Figure 6. The knotted cycle 13524 must be setwise invariant under every homeomorphism of Γ. Thus TSG₊(Γ) ≤ D₅. The automorphism (12345) is induced by rotating Γ, and (25)(34) is induced by turning the graph over. Hence TSG₊(Γ) = 〈(12345;, (25)(34)〉 ≅ D₅. Since the edge $\overline{12}$ is not pointwise fixed by any non-trivial element of TSG₊(Γ), by the Subgroup Theorem the groups ℤ₅ and ℤ₂ are also positively realizable for K₅.

Next consider the embedding Γ of K₅ illustrated in Figure 7. The triangle $\overline{123}$ must go to itself under any homeomorphism. Also by Lemma 1, any orientation preserving homeomorphism which fixes vertices 1, 2, and 3 induces a trivial automorphism on K₅. Thus TSG₊(Γ) ≤ D₃. The automorphism (123) is induced by a rotation. Also the automorphism (45)(12) is induced by pulling vertex 4 down through the centre of triangle $\overline{123}$while pulling vertex 5 into the centre of the figure then rotating by 180° about the line through vertex 3 and the midpoint of the edge $\overline{12}$. Thus TSG₊(Γ) = 〈(123;, (45)(12)〉 ≅ D₃. Since the edge $\overline{12}$is not pointwise fixed by any non-trivial element of TSG₊(Γ), by the Subgroup Theorem, the group ℤ₃ is positively realizable for K₅.

Lastly, consider the embedding Γ of K₅ illustrated in Figure 8 with vertex 5 at infinity. The square $\overline{1234}$must go to itself under any homeomorphism. Hence TSG₊(Γ) ≤ D₄. The automorphism (13)(24) is induced by rotating the square by 180°. By turning over the figure we obtain (12)(34). By Lemma 2, TSG₊(Γ) cannot contain an element of order 4. Thus TSG₊(Γ) = 〈(13)(24;, (12)(34)〉 ≅ D₂.

We summarize our results on positive realizability for K₅ in

Table 3. Non-trivial positively realizable groups for K₅.

Subgroup	Positively Realizable	Reason
S5	No	By Complete Graph Theorem
A5	Yes	By A5 Theorem
ℤ5 ⋊ ℤ4	No	By Complete Graph Theorem
S4	No	By S4 Theorem
A4	Yes	By A4 Theorem
D6	No	By Lemma 2
D5	Yes	By Figure 6
D4	No	By Lemma 2
D3	Yes	By Figure 7
D2	Yes	By Figure 8
ℤ6	No	By Lemma 2
ℤ5	Yes	By Subgroup Theorem
ℤ4	No	By Lemma 2
ℤ3	Yes	By Subgroup Theorem
ℤ2	Yes	By Subgroup Theorem

.

Again by adding appropriate equivalent chiral knots to each edge, all of the positively realizable groups for K₅ are also realizable. Thus we only need to determine realizability for the groups S₅, S₄, ℤ₅ ⋊ ℤ₄, D₆, D₄, ℤ₆,andℤ₄.

Let Γ be the embedding of K₅ illustrated in Figure 9. Any transposition which fixes vertex 5 is induced by a reflection through the plane containing the three vertices fixed by the transposition. To see that any transposition involving vertex 5 can be achieved, consider the automorphism (15). Pull $\overline{15}$through the triangle $\overline{234}$and then turn over the embedding so that vertex 5 is at the top, vertex 1 is in the centre and vertices 3 and 4 are switched. Now reflect in the plane containing vertices 1, 5, and 2 in order to switch vertices 3 and 4 back. All other transpositions involving vertex 5 can be induced by a similar sequence of moves. Hence TSG(Γ) ≅ S₅.

We create a new embedding Γ from Figure 9 by adding the achiral figure eight knot, 4₁, to all edges containing vertex 5. Now every homeomorphism of (S³, Γ) fixes vertex 5, yet all transpositions fixing vertex 5 are still possible. Thus TSG(Γ′) ≅ S₄.

In order to prove D₄ is realizable for K₅ consider the embedding Γ illustrated in Figure 10. Every homeomorphism of (S³, Γ) takes $\overline{1234}$to itself, so TSG(Γ) ≤ D₄. The automorphism (1234) is induced by rotating the graph by 90° about a vertical line through vertex 5, then reflecting in the plane containing the vertices 1, 2, 3, 4, and finally isotoping the knots into position. Furthermore, reflecting in the plane containing $\overline{153}$or $\overline{254}$and then isotoping the knots into position yields the transposition (24) or (13) respectively. Hence TSG(Γ) ≅ D₄.

We obtain a new embedding Γ by replacing the invertible 41 knots in Figure 10 with the knot 12₄₂₇, which is positive achiral but non-invertible [16]. Since 12₄₂₇ is neither negative achiral nor invertible, no homeomorphism of (S³, Γ) can invert $\overline{1234}$. Thus TSG(Γ′) ≅ ℤ₄.

Next let Γ denote the embedding of K₅ illustrated in Figure 11. Every homeomorphism of (S³, Γ) takes $\overline{123}$ to itself, so TSG(Γ) ≤ D₆. The 3-cycle (123) is induced by a rotation. Each transposition involving only vertices 1, 2, and 3 is induced by a reflection in the plane containing $\overline{45}$ and the remaining fixed vertex followed by an isotopy. The transposition (45) is induced by a reflection in the plane containing vertices 1, 2 and 3 followed by an isotopy. Thus TSG(Γ) ≅ D₆, generated by (123), (23), and (45).

We obtain a new embedding Γ′ by replacing the 4₁ knots in Figure 11 by 12₄₂₇ knots. Then the triangle $\overline{123}$ cannot be inverted. Thus TSG(Γ′) ≅ ℤ₆, generated by (123) and (45).

It is more difficult to show that ℤ₅ × ℤ₄ is realizable for K₅, so we define our embedding in two steps. First we create an embedding Γ of K₅ on a torus T that is standardly embedded in S³. In Figure 12, we illustrate Γ on a flat torus. Let f denote a glide rotation of S³ which rotates the torus longitudinally by 4π/5 while rotating it meridinally by 8π/5. Thus f takes Γ to itself inducing the automorphism (12345).

Let g denote the homeomorphism obtained by rotating S³ about a (1, 1) curve on the torus T, followed by a reflection through a sphere meeting T in two longitudes, and then a meridional rotation of T by 6π/5. In Figure 13, we illustrate the step-by-step action of g on T, showing that g takes Γ to itself inducing (2431).

The homeomorphisms f and g induce the automorphisms φ = (12345) and ψ = (2431) respectively. Observe that φ⁵ = ψ⁴ = 1 and ψφ = φψ². Thus 〈φ, ψ〉 ≅ ℤ₅ ⋊ ℤ₄ ≤ TSG(Γ) ≤ S₅. Note however that the embedding in Figure 12 is isotopic to the embedding of K₅ in Figure 9. Thus TSG(Γ) ≅ S₅.

In order to obtain the group ℤ₅ ⋊ ℤ₄, we now consider the embedding Γ of K₅ whose projection on a torus is illustrated in Figure 14. Observe that the projection of Γ′ in every square of the grid given by Γ on the torus is identical. Thus the homeomorphism f which took Γ to itself inducing the automorphism φ = (12345) on Γ also takes Γ′ to itself inducing φ on Γ′.

Recall that g was the homeomorphism of (S³, Γ) obtained by rotating S³ about a (1, 1) curve on the torus T, followed by a reflection through a sphere meeting T in two longitudes, and then a meridional rotation of T by 6π/5. In order to see what g does to Γ′, we focus on the square $\overline{1534}$ of Γ′. Figure 15 illustrates a rotation of the square $\overline{1534}$ about a diagonal, then a reflection of the square across a longitude. The result of these two actions takes the projection of the knot $\overline{1534}$ to an identical projection. Thus after rotating the torus meridionally by 6π/5, we see that g takes Γ′ to itself inducing the automorphism ψ = (2431). It now follows that ℤ₅ ⋊ ℤ₄ ≤ TSG(Γ′) < S₅.

In order to prove that TSG(Γ′) ≅ ℤ₅ ⋊ ℤ₄, we need to show TSG(Γ′) ≇ S₅. We prove this by showing that the automorphism (15) cannot be induced by a homeomorphism of (S³, Γ′).

From Figure 15 we see that the square $\overline{1534}$ is the knot 4₁#4₁#4₁#4₁. In order to see what would happen to this knot if the transposition (15) were induced by a homeomorphism of (S³, Γ), we consider the square $\overline{5123}$. In Figures 16 and 17 we isotop $\overline{5134}$ to a projection with only 10 crossings. This means that $\overline{5134}$ cannot be the knot 4₁#4₁#4₁#4₁. It follows that the automorphism (15) cannot be induced by a homeomorphism of (S³, Γ). Hence TSG(Γ′) ≇ S₅. However, the only subgroup of S₅ that contains ℤ₅ ⋊ ℤ₄ and is not S₅ is the group ℤ₅ ⋊ ℤ₄. Thus in fact TSG(Γ′) ≅ ℤ₅ ⋊ ℤ₄.

Thus every subgroup of Aut(K₅) is realizable for K₅.

Table 4. Non-trivial realizable groups for K₅.

Subgroup	Realizable	Reason
S5	Yes	By Figure 9
A5	Yes	Positively realizable
S4	Yes	By modifying Figure 9
A4	Yes	Positively realizable
D6	Yes	By Figure 11
D5	Yes	Positively realizable
D4	Yes	By Figure 10
D3	Yes	Positively realizable
D2	Yes	Positively realizable
ℤ6	Yes	By modifying Figure 11
ℤ5⋊ℤ4	Yes	By Figure 14
ℤ5	Yes	Positively realizable
ℤ4	Yes	By modifying Figure 10
ℤ3	Yes	Positively realizable
ℤ2	Yes	Positively realizable

summarizes our results for TSG(K₅).

6. Topological Symmetry Groups of K₆

The following is a complete list of all the subgroups of Aut(K₆) ≅ S₆: S₆, A₆, s₅, A₅, S₂[S₃] (B[A] represents a wreath product of A by B.), S₄ × ℤ₂, A₄ × ℤ₂, S₄, A₄, ℤ₅ ⋊ ℤ₄, D₃ × D₃, (ℤ₃ × ℤ₃) ⋊ ℤ₄, (ℤ₃ × ℤ₃) ⋊ ℤ₂, D₃ × ℤ₃, ℤ₃ × ℤ₃, D₆, D₅, D₄, D₄ × ℤ₂, D₃, D₂, ℤ₆, ℤ₅, ℤ₄, ℤ₄ × ℤ₂, ℤ₃, ℤ₂, ℤ₂ × ℤ₂ × ℤ₂ (see [17] and independently verified using the program GAP).

Consider the embedding Γ of K₆ illustrated in Figure 18. There are three paths in Γ on different levels that look like the letter “Z” which are highlighted in Figure 18. The top Z-path is $\overline{3146}$, the middle Z-path is $\overline{4251}$, and the bottom Z-path is $\overline{5362}$. The knotted cycle $\overline{123456}$ must be setwise invariant under every homeomorphism of Γ, and hence TSG₊(Γ) < D₆. The automorphism (123456) is induced by a glide rotation that cyclically permutes the Z-paths. Consider the homeomorphism obtained by rotating Γ by 180° about the line through vertices 2 and 5 and then pulling the edges $\overline{13}$ and $\overline{46}$ to the top level while pushing the lower edges down. The result of this homeomorphism is that the top Z-path $\overline{3146}$ goes to the top Z-path $\overline{1364}$, the middle Z-path $\overline{4251}$ goes to to middle Z-path $\overline{6253}$, and the bottom Z-path $\overline{5362}$ goes to the bottom Z-path $\overline{5142}$. Thus the homeomorphism leaves Γ setwise invariant inducing the automorphism (13)(46). It follows that TSG₊(Γ) = 〈(123456;, (13)(46)〉 ≅ D₆. Finally, since the edge $\overline{12}$ is not pointwise fixed by any non-trivial element of TSG₊(Γ), by the Subgroup Theorem the groups ℤ₆, D₃, ℤ₃, D₂ and ℤ₂ are positively realizable for K₆.

Consider the embedding, Γ of K₆ illustrated in Figure 19 with vertex 6 at infinity. The automorphisms (13524) and (25)(34) are induced by rotations. Also since 13524 is the only 5-cycle which is knotted, $\overline{13524}$ is setwise invariant under every homeomorphism of (S³, Γ). Hence TSG₊(Γ) ≅ D₅. Also since 15 is not pointwise fixed under any homeomorphism, by the Subgroup Theorem, ℤ₅ is positively realizable for K₆.

Next consider the embedding, Γ of K₆ illustrated in Figure 20. The automorphisms (123)(456) and (123)(465) are induced by glide rotations and (45)(12) is induced by turning the figure upside down. Also if we consider the circles $\overline{123}$ and $\overline{456}$ as cores of complementary solid tori, then (14)(25)(36) is induced by an orientation preserving homeomorphism that switches the two solid tori.

Observe that every homeomorphism of (S³, Γ) takes the pair of triangles $\overline{123}\cup \overline{456}$ to itself, since this is the only pair of complementary triangles not containing knots. The automorphism group of the union of two triangles is S₂[S₃] [18]. Thus TSG₊(Γ) ≤ S₂[S₃]. Note that the transpositions (12) and (45) are each induced by a reflection followed by an isotopy. Thus TSG(Γ) ≅ S₂[S₃], since (123)(456), (123)(465), (12) and (14)(25)(36) generate S₂[S₃]. However, by the Complete Graph Theorem, TSG₊(Γ) ≇ S₂[S₃]. Thus TSG₊(Γ) must be an index 2 subgroup of S₂[S₃] containing f = (123)(456), g = (123)(465), φ = (45)(12) and ψ = (14)(25)(36). Observe that φψ is the involution (42)(51)(36), and f commutes with ψ and also fφψ = φψf^-1, while g commutes with φψ and gψ = ψg^-1. Thus S₂[S₃] ≥ TSG₊(Γ) ≥ 〈f,φψ〉 × (g,ψ) ≅ D₃ × D₃. It follows that TSG+(Γ) ≅ D₃ × D₃.

The subgroup 〈f, g, ψ〉 is isomorphic to D₃ × ℤ₃ because ψ commutes with f and gψ = ψg^-1. We add the non-invertible knot 8₁₇ to every edge of the triangles $\overline{123}$ and $\overline{456}$ to obtain an embedding Γ₁. Now the automorphism φ = (45)(12) cannot be induced by an orientation preserving homeomorphism of (S³, Γ₁). However, f, g, and ψ are still induced by orientation preserving homeomorphisms. Thus TSG₊(Γ₁) ≅ D₃ × ℤ₃ since D₃ × ℤ₃ is a maximal subgroup of D₃ x D₃.

Also 〈f, g, φ〉 is isomorphic to (ℤ₃ × ℤ₃) ⋊ ℤ₂ because fφ = φf^-1 and gφ = φg^-1. Again starting with Γ in Figure 20, we place 5₂ knots on the edges of the triangle 123 so that ψ is no longer induced. Thus creating and embedding Γ₂ with TSG₊(Γ₂) ≅ (ℤ₃ × ℤ₃) ⋊ ℤ₂ since (ℤ₃ × ℤ₃) ⋊ ℤ₂ is a maximal subgroup of D₃ × D₃.

Finally 〈f, g〉 is isomorphic to ℤ₃ × ℤ₃. If we place equivalent non-invertible knots on each edge of the triangle $\overline{123}$ and a another set (distinct from the first set) of equivalent non-invertible knots on each edge of $\overline{456}$ we obtain an embedding Γ₃ with TSG₊(Γ₃) ≅ ℤ₃ × ℤ₃ since ℤ₃ × ℤ₃ is a maximal subgroup of (ℤ₃ × ℤ₃) ⋊ ℤ₂.

We summarize our results on positively realizability for K₆ in

Table 5. Non-trivial positively realizable groups for K₆.

Subgroup	Positively Realizable	Reason
A5	No	By A5 Theorem
S4	No	By S4 Theorem
A4	No	By A4 Theorem
D6	Yes	By Figure 18
D5	Yes	By Figure 19
D4	No	By Lemma 2
D3×D3	Yes	By Figure 20
D3×ℤ3	Yes	By modifying Figure 20
D3	Yes	By Subgroup Theorem
D2	Yes	By Subgroup Theorem
ℤ6	Yes	By Subgroup Theorem
ℤ5	Yes	By Subgroup Theorem
ℤ4	No	By Lemma 2
(ℤ3 × ℤ3) x ℤ2	Yes	By modifying Figure 20
ℤ3 ×ℤ3	Yes	By modifying Figure 20
ℤ3	Yes	By Subgroup Theorem
ℤ2	Yes	By Subgroup Theorem
S6,A 6,S5,S 2[S 3],S4×ℤ2,A4×ℤ2	No	By Complete Graph Theorem
ℤ5 ⋊ ℤ4, (ℤ3 × ℤ3) ⋊ ℤ4, D4 × ℤ2	No	By Complete Graph Theorem
ℤ4 ×ℤ2, ℤ2 ×ℤ2 ×ℤ2	No	By Complete Graph Theorem

. Note in the last few lines of the table we list multiple groups per line, since all of these groups are not positively realizable for the same reason.

By adding appropriate equivalent chiral knots to each edge, every group which is positively realizable for K₆ is also realizable for K₆. Thus we only need to determine realizability for the groups S₆, A₆, S₅, A₅, S₄ × ℤ₂, A₄ × ℤ₂, S₄, A₄, ℤ₅ ⋊ ℤ₄, (ℤ₃ × ℤ₃) ⋊ ℤ₄, D₄, D₄ × ℤ₂, ℤ₄, ℤ₄ × ℤ₂, and ℤ₂ × ℤ₂ × ℤ₂. Note that in Figure 20 we already determined that S₂[S₃] is realizable for K₆.

Let Γ₄ be the embedding of K₆ illustrated in Figure 20 with a left handed trefoil added to each edge of $\overline{123}$ and a right handed trefoil added to each edge of $\overline{456}$. The pair of triangles are setwise invariant since no other edges contain trefoils. Both (123)(456) and (123)(465) are induced by homeomorphisms of (Γ₄,S³). Also if we reflect in the plane containing vertices 4, 5, 6, and 1 then all the trefoils switch from left-handed to right-handed and vice versa. If we then interchange the complementary solid tori which have the triangles as cores followed by an isotopy, we obtain an orientation reversing homeomorphism that induces the order 4 automorphism (14)(25)(36)(23) = (14)(2536). Now 〈(14)(2536;, (123)(456), (123)(465)〉 ≅ (ℤ₃ × ℤ₃) ⋊ ℤ₄.

We see as follows that TSG(Γ₄) cannot be larger than (ℤ₃ × ℤ₃) ⋊ ℤ₄. Suppose that the automorphism (12) is induced by a homeomorphism f. By Lemma 1, f must be orientation reversing. But $f(\overline{456})=\overline{456}$, which is impossible because 456 contains only right handed trefoils. Thus TSG(Γ₄) ≇ S₂[S₃]. Note that the only proper subgroup of S₂[S₃] containing (ℤ₃ × ℤ₃) ⋊ ℤ₄ is (ℤ₃ × ℤ₃) ⋊ ℤ₄. Thus TSG(Γ₄) ≅ (ℤ₃ × ℤ₃) ⋊ ℤ₄.

Now let Γ be the embedding of K₆ illustrated in Figure 21. Observe that the linking number $\text{lk}(\overline{135},\overline{246})=\pm 1$, but $\text{lk}(\overline{136},\overline{245})=0$. Thus the automorphism (56) cannot be induced by a homeomorphism of (S³,Γ). Since every homeomorphism of (S³,Γ) takes $\overline{1234}$ to itself, it follows that TSG(Γ) ≤ D₄. The automorphism (1234)(56) is induced by a rotation followed by a reflection and an isotopy. In addition the automorphism (14)(23)(56) is induced by turning the figure upside down. Thus TSG(Γ) ≅ D₄ generated by the automorphisms (1234)(56) and (14)(23)(56).

Now let Γ be obtained from Figure 21 by replacing the knot 4₁ with the non-invertible and positively achiral knot 12₄₂₇. Then the square $\overline{1234}$ can no longer be inverted. In this case (1234)(56) generates TSG(Γ′) and thus TSG(Γ′) ≅ ℤ ₄.

For the next few groups we will use the following lemma.

4-Cycle Theorem

[19] For any embedding Γ of K₆ in S³, and any labelling of the vertices of K₆ by the numbers 1 through 6, there is no homeomorphism of (S³,Γ) which induces the automorphism (1234).

Consider the subgroup ℤ₅ ⋊ ℤ₄ ≤ Aut(K₆). The presentation of ℤ₅ ⋊ ℤ₄ as a subgroup of S₆ gives the relation x^-1yx = y² for some elements x, y ∈ ℤ₅ ⋊ ℤ₄ of orders 4 and 5 respectively. Suppose that for some embedding Γ of K₆, we have TSG(Γ) ≅ ℤ₅ ⋊ ℤ₄. Without loss of generality, we can assume that y = (12345) satisfies the relation x^-1yx = y². By the 4-Cycle Theorem, any order 4 element of TSG(Γ) must be of the form x = (abcd)(ef). However, there is no element in Aut(K₆) of the form x = (abcd)(ef) that together with y = (12345) satisfies this relation. Thus there can be no embedding Γ of K₆ in S³ such that TSG(Γ) ≅ ℤ₅ ⋊ ℤ₄.

Now consider the subgroup ℤ₄ × ℤ₂ ≤ Aut(K₆). By the 4-Cycle Theorem, without loss of generality we may assume that if TSG(Γ) contains an element of order 4 for some embedding Γ of K₆, then TSG(Γ) contains the element (1234)(56). Computation shows that the only transposition in Aut(K₆) that commutes with (1234)(56) is (56), which cannot be an element of TSG(Γ) since this would imply that (1234) is an element of TSG(Γ). Furthermore the only order 2 element of Aut(K₆) that commutes with (1234)(56) and is not a transposition is (13)(24), which is already in the group generated by (1234)(56). Thus there is no embedding Γ of K₆ in S³ such that TSG(Γ) contains the group ℤ₄ × ℤ₂. This rules out all of the groups S₄ × ℤ₂, D₄ × ℤ₂ and ℤ₄ × ℤ₂ as possible topological symmetry groups for embeddings of K₆ in S³.

For the group ℤ₂ × ℤ₂ × ℤ₂ we will use the following result.

Conway Gordon

[20] For any embedding Γ of K₆ in S³, the mod 2 sum of the linking numbers of all pairs of complementary triangles in Γ is 1.

Now suppose that for some embedding Γ of K₆ in S³ we have TSG(Γ) ≅ ℤ₂ × ℤ₂ × ℤ₂. It can be shown that the subgroup ℤ₂ × ℤ₂ × ℤ₂ ≤ Aut(K₆) contains three disjoint transpositions. Without loss of generality we can assume that TSG(Γ) contains (13), (24), and (56), which are induced by homeomorphisms h, f, and g of (S³, Γ) respectively. Since any three vertices of Γ determine a pair of disjoint triangles, we can use a triple of vertices to represent a pair of disjoint triangles. For example, we use the triple 123 to denote the pair of triangles $\overline{123}$ and $\overline{456}$. With this notation, the orbits of the ten pairs of disjoint triangles in K₆ under the group 〈(13;, (24), (56)〉 are:

$$\{124,143\},\{124,324\},\{125,325,145,126\},\{135,136\}$$

Since h, f, and g are homeomorphisms of (S³, Γ) the links in a given orbit all have the same (mod 2) linking number. Since each of these orbits has an even number of pair of triangles, this contradicts Conway Gordon. Thus ℤ₂ × ℤ₂ × ℤ₂ ≇ TSG(Γ). Hence ℤ₂ × ℤ₂ × ℤ₂ is not realizable for K₆

Table 6. Non-trivial realizable groups for K₆.

Subgroup	Realizable	Reason
S6	No	TSG+(K6) cannot be S6 or A6
A6	No	TSG+(K6) cannot be A6
S5	No	TSG+(K6) cannot be S5 or A5
A5	No	TSG+(K6) cannot be A5
S4 × ℤ2	No	TSG+(K6) cannot be S4 × ℤ2 or S4
S4	No	TSG+(K6) cannot be S4 or A4
A4 × ℤ2	No	TSG+(K6)cannotbeA4 × ℤ2orA4
A4	No	TSG+(K6) cannot be A4
D6	Yes	Positively realizable
D5	Yes	Positively realizable
D4×ℤ2	No	TSG+(K6) cannot be D4 × ℤ2, D4, ℤ4 × ℤ2, ℤ2 × ℤ2 × ℤ2
D4	Yes	By Figure 21
S2[S3]	Yes	By Figure 20
D3 × D3	Yes	Positively realizable
D3 × ℤ3	Yes	Positively realizable
D3	Yes	Positively realizable
D2	Yes	Positively realizable
ℤ6	Yes	Positively realizable
ℤ5×ℤ4	No	By 4-Cycle Theorem
ℤ5	Yes	Positively realizable
ℤ4×ℤ2	No	By 4-Cycle Theorem
ℤ4	Yes	By modifying Figure 21
(ℤ3 × ℤ3) ⋊ ℤ4	Yes	By modifying Figure 20
(ℤ3 × ℤ3) ⋊ ℤ2	Yes	Positively realizable
ℤ3×ℤ3	Yes	Positively realizable
ℤ3	Yes	Positively realizable
ℤ2 x ℤ2 × ℤ2	No	By Conway Gordon Theorem
ℤ2	Yes	Positively realizable

summarizes our realizability results for K₆. Recall that for n = 4 and n = 5 every subgroup of S_n is realizable for K_n. However, as we see from Table 6, this is not true for n = 6.

7. Conclusions

We have classified all groups which can occur as the topological symmetry group or orientation preserving topological symmetry group of an embedded complete graph with no more than six vertices. Our results show that a number of groups can occur as a topological symmetry group but not as an orientation preserving topological symmetry group for a particular K_n. This gives us a collection of groups which can only occur for achiral embeddings of the graph in question.

The topological symmetry group includes all of the symmetries induced by the point group together with any symmetries that occur as the result of any flexibility or rotation of subparts of a structure around specific bonds. Thus the topological symmetry group gives us more information about the symmetries and possible achirality of supramolecular structures than could be obtained from the point group. Since complete graphs with no more than six vertices may occur as supramolecular clusters, these results could be of interest in the future study of supramolecular chirality.